For any set $X$, let $\newcommand{\N}{\mathbb{N}}[X]^2 = \big\{\{x,y\}:x\neq y \in X\big\}$ and set $[n]^2 = [\{0,\ldots,n-1\}]^2$ for any positive integer $n$. For $A\subseteq [\N]^2$ we set $$\newcommand{\mmu}{\mu_{[\N]^2}}\mmu(A) = \lim\inf_{n\to\infty}\frac{ \text{card}(A\cap [n+1]^2)}{\text{card}([n+1]^2)}.$$

If $f:\N\to\N$ is a bijection, we set $$\newcommand{\rev}{\text{rev}}\rev(f) =\{p\in[\N]^2:\min(f(p)) = f(\max(p))\}.$$

Is there a bijection $f:\N\to\N$ with $\mmu\big(\rev(f)\big) = 1$?

For any set $X$, let $\newcommand{\N}{\mathbb{N}}[X]^2 = \big\{\{x,y\}:x\neq y \in X\big\}$ and set $[n]^2 = [\{0,\dotsc,n-1\}]^2$ for any positive integer $n$. For $A\subseteq [\N]^2$ we set $$\newcommand{\mmu}{\mu_{[\N]^2}}\mmu(A) = \liminf_{n\to\infty}\frac{ \operatorname{card}(A\cap [n+1]^2)}{\operatorname{card}([n+1]^2)}.$$

If $f:\N\to\N$ is a bijection, we set $$\DeclareMathOperator{\rev}{rev}\rev(f) =\{p\in[\N]^2:\min(f(p)) = f(\max(p))\}.$$

Is there a bijection $f:\N\to\N$ with $\mmu\big({\rev(f)}\big) = 1$?